For an irreducible smooth (generic) representation $\pi$ of $G=GL_2(k)$ with central character $\omega$, where $k$ is a $p$-adic field, we define the conductor of a vector $v\in\pi$ as follows. Let $K_0({\mathfrak p}^N)$ be the subgroup of $K=GL_2({\mathfrak o})$ with lower left-hand entry congruent to $0$ modulo ${\mathfrak p}^N$. The conductor of $v$, $c(v)$, is the smallest $N$ such that
$$\pi\bigg(\matrix{a&b\cr c&d}\bigg)v=\omega(a)v\ \ {\rm for\ all}\ \bigg(\matrix{a&b\cr c&d}\bigg)\in K_0({\mathfrak p}^N)$$
In other words, $v$ is fixed by $K_0({\mathfrak p}^N)$ up to the action of the center. The conductor of $\pi$, $c(\pi)$, is the smallest $N$ such that there is a $v$ with $c(v)=N$. The subspace of vectors with conductor $c(\pi)$ is one-dimensional and the unique vector $v_0$ that maps to a Whittaker function with $W_0(1)=1$ is called the new vector.

My question is: how can we write down a somewhat-explicit basis of $\pi$ in terms of the new vector $v_0$, indexed by the principal congruence subgroups $K({\mathfrak p}^M)=1_2+{\mathfrak p}^MM_2({\mathfrak o})$? In other words, how can we write a basis of $\pi^{K({\mathfrak p}^M)}$ in terms of $v_0$?

A simple nonexplicit version is this result, from Casselman's "The restriction of a representation of $GL_2(k)$ to $GL_2({\mathfrak o})$": the restriction of $\pi$ to $K$ decomposes as
$$res_K^G\pi=\pi^{K({\mathfrak p}^{c(\pi)-1})}\oplus\sum_{n\ge c(\pi)}u_n(\omega)$$
where $u_n$ is the unique irreducible representation of $K$ which is trivial on $K({\mathfrak p}^n)$ but not on $K({\mathfrak p}^{n-1})$ and that contains a vector $v$ with conductor $n$ in the above sense.

Btw, I think there is a small mistake: if $c(\pi)$ is not zero, all $K$-reps are higher dimensional, so the $K(\pi)$-invariant vectors are not one-dimensional. But all $K$-isotypes occur with multiplicity one, which is essential for giving an easy proof for the commutativity of the (ramified) Hecke operators.
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Marc PalmOct 15 '12 at 9:55

Hi Marc, I think you might have misread, though maybe I'm missing something: the one-dimensional space of vectors is fixed by $K_0(p^N)$ rather than $K(p^N)$ (i.e. they are newforms).
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B ROct 22 '12 at 16:33

1 Answer
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I think that your asking how to map from one isotype to another, right?

Now, one way to map from one isotype to an other is convolution with a function
$$\phi = \langle v_1, f v_2\rangle, \qquad f(k_1gk_2) =u_{n_1}(k_1) f(g) u_{n_2}(k_2),$$
where $g$ runs through a set of representatives of $G//K$, and $f$ is $Hom_{\mathbb{C}}(u_{n_2},u_{n_1})$-valued. $v_1$ and $v_2$ are vectors. This follows from Schur's orthogonality relations.

There is possibe a nicer way to express this, but this is essentially the only way! For what purpose do you need this?

Hi Marc, sorry for taking so long to respond! I am exactly trying to move through the types. In fact, Reeder's Oldforms paper does mostly what I want to do. This is related to the other question of mine you answered: I'm doing a calculation that runs through a basis of a representation and needs to be compatible with the $K$-type structure, and I'm trying to write it in a way that I can work with.
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B ROct 22 '12 at 16:42

Okay, and I was suggesting a general (though not very encouraging) method. You could start with highest level with the vector $f(bk')$, where $b \in B$ and $k' \in K_0(p^N)$, and the write $\sigma_{N+1}(\mu)$ as the orthogonal difference of induced representation, compute a matrix coefficient, convolute and so forth... But I see now that you want something less messy and elaborate. I have done this one time for the Steinberg, and this is already a mess:( For the remaining principal series representation, I have kept only working with th K-types associated with fundamental type....
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Marc PalmOct 22 '12 at 17:09

i.e., $\sigma_N(\mu)$, since going to lower level types, e.g. $\sigma_{N+1}(\mu)$ mixes only automorphic representation with different principal series representation at $p$ (i.e. those $\mu$'s coinciding on $1+p^{N+1}$). I like to consult only principal series with varying spectral character, but the same $B$-rep. Am I still clear?
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Marc PalmOct 22 '12 at 17:13